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Daily Sudoku — 20 July 2026

The puzzle

  • 20 July 2026
  • Medium
  • 34 clues
Puzzle
·2·····64
4··2··1··
·135··279
··9··1··5
·34····8·
·78·4·6·2
····5··2·
·8··26917
···9···53
Solution
927138564
456279138
813564279
269781345
534692781
178345692
791853426
385426917
642917853

Solve it in the app to get on the board

Step-by-step solution

Every cell below is filled by the same hint engine the app uses, in the order it can be deduced — 47 steps using 2 techniques.

Techniques used

  • Hidden Single — inside one row, column or box this digit fits in only one cell, so it must go there
  • Naked Single — every other digit is already used in this cell's row, column or box, so only one candidate is left
  1. R3C6 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C4, R1C5 and R1C6 cannot hold 4.- Because row 2 already has a 4, R2C5 and R2C6 cannot hold 4.- Because column 5 already has a 4, R3C5 cannot hold 4.So 4 must go in R3C6.
  2. R1C7 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 5, R2C8 cannot hold 5.- Because column 9 already has a 5, R2C9 cannot hold 5.So 5 must go in R1C7.
  3. R2C8 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R2C9 cannot hold 3.So 3 must go in R2C8.
  4. R2C9 = 8 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3 and 4 → ruled out.- Column 9 already has 5, 7 and 9 → ruled out.- The top-right 3×3 box already has 6 → ruled out.So this cell can only be 8.
  5. R5C6 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 2, R4C4 and R5C4 cannot hold 2.- Because column 5 already has a 2, R4C5 and R5C5 cannot hold 2.- Because row 6 already has a 2, R6C4 and R6C6 cannot hold 2.So 2 must go in R5C6.
  6. R4C1 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R4C2 cannot hold 2.- Because row 5 already has a 2, R5C1 cannot hold 2.- Because row 6 already has a 2, R6C1 cannot hold 2.So 2 must go in R4C1.
  7. R6C6 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C4 and R4C5 cannot hold 5.- Because column 4 already has a 5, R5C4 and R6C4 cannot hold 5.- Because column 5 already has a 5, R5C5 cannot hold 5.So 5 must go in R6C6.
  8. R5C1 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C2 cannot hold 5.- Because row 6 already has a 5, R6C1 cannot hold 5.So 5 must go in R5C1.
  9. R6C1 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C2 cannot hold 1.So 1 must go in R6C1.
  10. R4C2 = 6 · Naked SingleLooking at R4C2:- Row 4 already has 1, 2, 5 and 9 → ruled out.- Column 2 already has 3, 7 and 8 → ruled out.- The middle-left 3×3 box already has 4 → ruled out.So this cell can only be 6.
  11. R5C5 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C4 and R4C5 cannot hold 9.- Because column 4 already has a 9, R5C4 and R6C4 cannot hold 9.So 9 must go in R5C5.
  12. R5C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C4 and R4C5 cannot hold 6.- Because row 6 already has a 6, R6C4 cannot hold 6.So 6 must go in R5C4.
  13. R5C9 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C7 and R4C8 cannot hold 1.- Because column 7 already has a 1, R5C7 cannot hold 1.- Because row 6 already has a 1, R6C8 cannot hold 1.So 1 must go in R5C9.
  14. R5C7 = 7 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  15. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 2 and 5 → ruled out.- Column 9 already has 1, 3, 4, 7, 8 and 9 → ruled out.So this cell can only be 6.
  16. R4C7 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 3, R4C8 and R6C8 cannot hold 3.So 3 must go in R4C7.
  17. R6C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C4 and R4C5 cannot hold 3.So 3 must go in R6C4.
  18. R6C8 = 9 · Naked SingleLooking at R6C8:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  19. R4C8 = 4 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 5, 6 and 9 → ruled out.- Column 8 already has 7 and 8 → ruled out.So this cell can only be 4.
  20. R9C3 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C1, R7C2 and R7C3 cannot hold 2.- Because row 8 already has a 2, R8C1 and R8C3 cannot hold 2.- Because column 1 already has a 2, R9C1 cannot hold 2.- Because column 2 already has a 2, R9C2 cannot hold 2.So 2 must go in R9C3.
  21. R7C3 = 1 · Hidden SingleIn the bottom-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 1, R7C1 and R9C1 cannot hold 1.- Because column 2 already has a 1, R7C2 and R9C2 cannot hold 1.- Because row 8 already has a 1, R8C1 and R8C3 cannot hold 1.So 1 must go in R7C3.
  22. R8C3 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C1 and R7C2 cannot hold 5.- Because column 1 already has a 5, R8C1 cannot hold 5.- Because row 9 already has a 5, R9C1 and R9C2 cannot hold 5.So 5 must go in R8C3.
  23. R2C2 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C1 and R1C3 cannot hold 5.- Because column 3 already has a 5, R2C3 cannot hold 5.- Because row 3 already has a 5, R3C1 cannot hold 5.So 5 must go in R2C2.
  24. R1C1 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R1C3 and R2C3 cannot hold 9.- Because row 3 already has a 9, R3C1 cannot hold 9.So 9 must go in R1C1.
  25. R3C1 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 8, R1C3 cannot hold 8.- Because row 2 already has a 8, R2C3 cannot hold 8.So 8 must go in R3C1.
  26. R3C5 = 6 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  27. R2C3 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C3 cannot hold 6.So 6 must go in R2C3.
  28. R1C3 = 7 · Naked SingleLooking at R1C3:- Row 1 already has 2, 4, 5, 6 and 9 → ruled out.- Column 3 already has 1, 3 and 8 → ruled out.So this cell can only be 7.
  29. R2C6 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C4, R1C5 and R1C6 cannot hold 9.- Because column 5 already has a 9, R2C5 cannot hold 9.So 9 must go in R2C6.
  30. R2C5 = 7 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  31. R4C4 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R4C5 cannot hold 7.So 7 must go in R4C4.
  32. R4C5 = 8 · Naked SingleLooking at R4C5:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  33. R9C1 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C1 and R7C2 cannot hold 6.- Because row 8 already has a 6, R8C1 cannot hold 6.- Because column 2 already has a 6, R9C2 cannot hold 6.So 6 must go in R9C1.
  34. R7C1 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 7, R7C2 and R9C2 cannot hold 7.- Because row 8 already has a 7, R8C1 cannot hold 7.So 7 must go in R7C1.
  35. R8C1 = 3 · Naked SingleLooking at R8C1:- Row 8 already has 1, 2, 5, 6, 7, 8 and 9 → ruled out.- Column 1 already has 4 → ruled out.So this cell can only be 3.
  36. R8C4 = 4 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  37. R7C2 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 9, R9C2 cannot hold 9.So 9 must go in R7C2.
  38. R9C2 = 4 · Naked SingleLooking at R9C2:- Row 9 already has 2, 3, 5, 6 and 9 → ruled out.- Column 2 already has 1, 7 and 8 → ruled out.So this cell can only be 4.
  39. R9C5 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C4 and R7C6 cannot hold 1.- Because column 6 already has a 1, R9C6 cannot hold 1.So 1 must go in R9C5.
  40. R1C5 = 3 · Naked SingleLooking at R1C5:- Row 1 already has 2, 4, 5, 6, 7 and 9 → ruled out.- Column 5 already has 1 and 8 → ruled out.So this cell can only be 3.
  41. R1C4 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 1, R1C6 cannot hold 1.So 1 must go in R1C4.
  42. R1C6 = 8 · Naked SingleLooking at R1C6:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  43. R7C4 = 8 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 5, 6, 7 and 9 → ruled out.- Column 4 already has 3 and 4 → ruled out.So this cell can only be 8.
  44. R7C6 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 3, R9C6 cannot hold 3.So 3 must go in R7C6.
  45. R7C7 = 4 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  46. R9C6 = 7 · Naked SingleLooking at R9C6:- Row 9 already has 1, 2, 3, 4, 5, 6 and 9 → ruled out.- Column 6 already has 8 → ruled out.So this cell can only be 7.
  47. R9C7 = 8 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.

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